What is [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)] ?

The solution to [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)] is 0

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simplify [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)]

simplify

[3 ln (3) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

0

Solution steps

[3 ln (3) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 ln (3) = ln (3^{3})

= [ln (3^{3}) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (2) = ln (2^{2})

= [ln (3^{3}) - 1 - ln (2^{2})] - [3 ln (3) - 1 - 2 ln (2)]

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (3^{3}) - ln (2^{2}) = ln (\frac{3 ^{3}}{2 ^{2}})

= [ln (\frac{3 ^{3}}{2 ^{2}}) - 1] - [3 ln (3) - 1 - 2 ln (2)]

\frac{3 ^{3}}{2 ^{2}} = \frac{27}{4}

= [ln (\frac{27}{4}) - 1] - [3 ln (3) - 1 - 2 ln (2)]

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 ln (3) = ln (3^{3})

= [ln (\frac{27}{4}) - 1] - [ln (3^{3}) - 1 - 2 ln (2)]

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (2) = ln (2^{2})

= [ln (\frac{27}{4}) - 1] - [ln (3^{3}) - 1 - ln (2^{2})]

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (3^{3}) - ln (2^{2}) = ln (\frac{3 ^{3}}{2 ^{2}})

= [ln (\frac{27}{4}) - 1] - [ln (\frac{3 ^{3}}{2 ^{2}}) - 1]

\frac{3 ^{3}}{2 ^{2}} = \frac{27}{4}

= [ln (\frac{27}{4}) - 1] - [ln (\frac{27}{4}) - 1]

Remove parentheses:

(a) = a

= ln (\frac{27}{4}) - 1 - (ln (\frac{27}{4}) - 1)

- (ln (\frac{27}{4}) - 1) : - ln (\frac{27}{4}) + 1

= ln (\frac{27}{4}) - 1 - ln (\frac{27}{4}) + 1

Simplify

ln (\frac{27}{4}) - 1 - ln (\frac{27}{4}) + 1 : 0

= 0

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ln (x y) + ln (\frac{x}{e})

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What is [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)] ?
The solution to [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)] is 0